CHAPTER 2: DISTILLATION PROCESS

Subtopics

Vapor-Liquid Equilibrium Types of Distillation Mass Balance in a Distillation

Column Determination of Ideal Number

of Plates using Mc-Cabe Thiele Method

Multi components Distillation

Distillation is a process wherein a liquid or vapour mixture of two or more substances is separated into its component fractions of desired purity, by the application and removal of heat.

method of separating mixtures based on differences in their volatilities in a boiling liquid mixture.

less volatile, "heavy" or "high boiling point", components concentrate in the liquid phase; the more volatile, "light“ or “low boiling point”, components concentrate in the vapor.

used for many commercial processes, such as production of gasoline, distilled water, alcohol, and many other liquids

What is Distillation?

Introduction

Vapor-Liquid Equilibrium (VLE)

Equilibrium in Chemical Engineering

Chemical equilibrium rates of reaction in both directions are same.

Phase equilibrium the rate of changing from one phase to another is same to the rate of the reverse change.

Vapor-liquid phase equilibrium: ????

Condition or state where the rate of evaporation (liquid changing to vapor) equals the rate of condensation (vapor changing to liquid).

VLE data can be determined experimentally using an equilibrium still.

VLE data can be determined or approximated with the help of certain theories such as Raoult's Law, Dalton's Law, and/or Henry's Law.

1. Pxy diagram: x and y as functions of pressure at constant temperature.

2. Txy diagram: x and y as functions of temperature at constant pressure.

3. xy diagram: x versus y at constant pressure (temperature is a parameter along the curve).

Since most applications require data at constant pressure, Txy and xy diagrams are the most commonly used.

Binary system-VLE Data

There are several different types of plots for binary system:

Txy Diagram (Phase Diagram)

xy diagram for a binary system,

relates the compositions of the liquid

and vapor phases in equilibrium with

each other.

xy Diagrams

These diagrams be generated from

Constant pressure- boiling point

diagram

xy diagram for binary system

Binary system-VLE Data

Binary system-VLE Data

How to present

VLE data?

Temperature-composition diagram(Txy)

T

xA

Tb(B)

Tb(A)

V

L

T1

T2

T3

T4

x1 y1x2 y2x3 y3x4 y4

Step 1

Binary system-VLE DataVLE data is obtained from Boiling points diagram

Binary system-VLE DataStep 3 Plots x-y diagram

yA

xA

T1

T2

T3

T4

Raoult’s Law

Where

pi= partial pressure of species i in the vapor

Pi o = the vapor pressure of pure species

xi=mole fraction of species i in the liquid

12

VLE RelationshipIf experimental data are not available, estimation of VLE can still be done. HOW?

simplest method assumes ideal vapor and ideal liquid phases.

isatii xPPy i

oii xPp

Calculations Using Raoult's Law

Bubble-point pressure problem -- T,x given -- P,y unknown.

The vapor pressures are found at the given temperature, which allows direct calculation of the pressure and vapor mole fractions:

P

xPy

xPPPy

isati

i

isatii

Where

P = total pressure of component A in the vapor.

= vapor pressure of species i

xi =mole fraction of species i in the liquid

satiP

VLE at Low Pressures – Raoult’s Law

Calculations Using Raoult's Law

Dew-point pressure problem -- T,y given -- P,x unknown.

No trial and error is needed, as P can be directly calculated.

xPy

Px

Py P

ii

isat i

i isat

1

1

/

VLE at Low Pressures – Raoult’s Law

Example 1: (Use of Raoult’s Law for boiling point Diagram)

Use Raoult's Law and calculate the vapour and liquid compositions in equilibrium at 95Co (368.2 K) (in mole fractions, y and x) for the benzene-toluene system using vapour pressure data measure at a pressure of 101.32 kPa as shown in Table 1 below :

Table 1:

from Table 1

For benzene P0A= 155.7 kPa

For toluene P0B= 63.3 kPa

Refer to equation Dalton Law:

155.7(xA) + 63.3 (1-xA)=101.32 kPa

Liquid composition;

xA=0.411 and

xB= 1-0.411=0.589 kPa

PxPxP AoBA

oA )1(

P

xPy A

oA

A

Vapor composition:

yA= (155.7 x 0.411)/101.32 = 0.632yB= 1-0.632=0.368

Raoult’s Law

pi=Pi o

xi

Dalton’s Law

Ppp BA

xy data

Relative Volatility of Vapor-Liquid Systems

)1)(1(

/

/

/

AA

AA

BB

AAAB xy

xy

xy

xy

AAB

AABA x

xy

)1(1

P

xPy AAA

0

0

0

B

AAB P

P

P

xPy BBB

0

Relative volatility

It is a measure of the differences in volatility between 2 components, and hence their boiling points. It indicates how easy or difficult a particular separation will be.

Where αAB is the relative volatility of A with respect to B in the binary system.

when αAB is above 1.0, a separation is possible.

Raoult’s law:

)( AB

Example: Using data from table 1 calculate the relative volatility for the benzene-toluene system at 85ºC (358.2K) and 105ºC (378.2K)

Solution: At 85ºC, substituting into equation below for a system following Raoult’s law,

Similarly at 105ºC,

54.20.46

9.1160

0

B

AAB P

P

38.20.86

2.204

Types of distillation column

There are 3 types in which the distillation may be carried out

Differential or batch distillation

Flash Distillation or equilibrium distillation

Continuous Distillation with reflux – Binary systems

INDIVIDUAL ASSIGNMENT

provide the necessary vaporization for the distillation process

To cool and condense the vapor leaving the top of the column

To hold the condensed vapor so that liquid (reflux) can be recycled back to the column

Trays/plates and/or packings which are used to enhance component separations

Bottoms B -richer in the less volatile component, where the mole fraction of the more volatile component is, xB

Distillate D which is richer in the more volatile component of mole fraction, xD.

Distillation with reflux and McCABE-THIELE method

Rectification (fractionation )or stage distillation with reflux ;

can be considered to be a process in which a series of flash-vaporization stages are arranged in a series in such a manner that the vapor and liquid products from each stage flow counter current to each other

Hence in each stage , a vapor V and a liquid stream L enter, are contact and mixed and equilibrated , and a vapor and a liquid stream leave in equilibrium

At each stage of the column two phases come in contact with each other, mix, approach thermal and composition equilibrium to the extent which depends on the efficiency of the contact stage

Lin,xin

Lout,xout

Vout,yout

Vin,yin

Streams leaving the stage are in thermodynamicequilibrium with each other

Streams coming to the stage are not in equilibrium

McCabe-Thiele method of calculation for Number of theoretical Stages

It is a mathematical graphical method for determining the number of theoretical trays or stages needed for a given separation of a binary mixture of A and B.

The main assumption in this method is that There must be an equimolar flow through the tower between the feed inlet and the top tray and the feed inlet and the bottom tray.

Action on an Ideal Plate By definition, a vapour and liquid leaving a plate are

brought into equilibrium. Assume that the plates are numbered serially from

top down and that the plate under consideration is the nth plate from the top.

Then the immediately above plate n is plate n-1, and the immediately below is n+1.

Ln-1

Xn-1

Plate n-1

Plate n

Plate n+1

Vn-1

yn-1

Ln,

xn

Ln+1

Xn+1

Vn+1

yn+1

Vn+2,

yn+2

V n

y n

Ln-2

Xn-2

Material –balance diagram for plate n

Material balances for two components systems

1. Total material balance on the entire column

F = D + B

2. Component material balance on component A

F xF= D xD+ BxB

Fig 1.10: Material balance for continuous fractionating column

W=B

Material Balances (top section)

Material balance around condenser:

Overall material balance over the Fig 1.11:

Components material balance over the Fig 1.11:

28

DLV nn 1

DLV

D

LnR

where

xR

xR

Ry

xV

Dx

V

Ly

D

Dn

nn

nn

1

1

1

xD xL yV

111

Dnn1n1n

Reflux ratio = constant

29

11

R

R

V

LR

slopeD

LnR

n

n

Dnn xR

xR

Ry

1

1

11

y

xxD

slope=R/(R+1)

DxR 1

1

Operating Line: Rectifying

30

Material Balances (bottom section- Stripping)

111

11

1

m

wm

m

mm

Wmmmm

mm

V

Wxx

V

Ly

WxxLyV

WLV

1m

m

V

L

Since equimolar flow is assumed ,the slop is

liquid flow to plate m+1 = Vapour flow from plate m+1 + Bottom product withdrawn

Overall components material balance over plate m+1:

Rearranging the equation :

Feed Line

The conditions of the vapor rate or the liquid rate may change depending of the thermal condition of the feed.

It is related to the heat to vaporize one mole of feed divided by molar latent heat (q)

It is the locus of the intersection of the two operating lines

Its intersection with the 450 line is y=x=xf where xf is the overall composition of the feed.

Feed Line Equation

If xq = xF, and yq =xF then;

The point of intersection of the two operating lines lies on the straight line of slope (q/q -1) and intercept (xF, yF)

1

the heat needed to vaporize mole of feed entering conditionsq

molar latent heat of vaporization of feed

11

q

xx

q

qy F

qq

y=x

x=xf

q<0

q=0

q=1 q>1y

x

11

q

xx

q

qy F

qq

Cold feed Feed at saturated liquid

Feed at saturated vapour

Feed superheated

Feed line behavior

35

Feed partial vapor 0<q<1

THEORETICAL STAGES

Starting at xD and stepping of the plate xW

Since reboiler is considered a theoretical step, the no of theoretical trays in a tower is equal to the number of the theoretical step, minus 1.

No of trays = No of steps– 1(reboiler)

4 stages + reboiler

Construction for the McCabe-Thiele Method

45° line

x=zFxB

y

xD

Step 1: Plot equilibrium curve and 45 degree line.Step 2: Plot given compositions (xF, xB, and xD)Step 3: Draw q-line from xF and yF

Step 4: Determine Rmin from intersection of therectifying section OL and the equilibrium curve.Step 5: Determine R from R/Rmin

Step 6: Draw OL for Rectifying sectionStep 7: Draw OL for Stripping section

equilibrium curve

x

45° liney

equilibrium curve

x=zFxB xD

y

equilibrium curve

x=zFxB xD

y

equilibrium curve

R/(R+1)

x=zFxB xD

y

equilibrium curve

x=zFxB xD

y

equilibrium curve

Rmin/(Rmin+1)

1. 2.

3. 4. 5. and 6. 7.

Dnn xR

xR

Ry

1

1

11

q

xx

q

qy F

11

y

x

zf

zf

xB xD

y1

yB

xN

DxR 1

1

Complete picture McCabe Thiele

111

m

wm

m

mm V

Wxx

V

Ly

y

x

zf

zf

xB xD

y1

yB

xN

Complete picture McCabe ThieleStep 1: Plot equilibrium curve(VLE) data.Step 2: Plot 45 degree line(diagonal line. y=x)Step 3: Plot given compositions (xF, xB, and xD)Step 4: Draw q-line from xF and yF

Step 5: Draw OL for Rectifying sectionStep 6 : Draw OL for Stripping sectionStep 7: Start stepping off from the distillate end until

the intersection of the two operating lines is passed.

Step 8: Continue stepping but use the stripping operating line.

Step 9: Count the number of stages.Step 10: Subtract one for the reboiler to give

the number of theoretical trays

Reflux Ratio

The analysis of fractionating columns is facilitated by the use of a quantity called reflux ratio.

Two ratios are used, one is the ratio of the reflux to the overhead product and the other is the ratio of the reflux to the vapor.

Both ratios refer to quantities in the rectifying section. The equations for those ratios are

DL

L

V

LRand

D

DV

D

LR VD

Minimum Reflux Ratio Rm

• Reflux ratio, R that will require an infinite number of plate for the given desired separation of xD and xB

• At any reflux less than total, the number of plates needed is larger than at total reflux and increases continuously as the reflux ratio decreased.

• This corresponds to the minimum amount of liquid return in the tower, and hence the minimum reboiler duty and condenser cooling capacity

If R is decreased, the slope of the (ROL) operating line R/(R + 1) is decreased, and the intersection of this line and the stripping line with the q line moves farther from the 450 line and closer to the equilibrium line.

To achieve separation, the number of steps required to give a fixed xD and xW increases. Separation more difficult

when driving force of mass transfer is zero (operation at equilibrium point)

Minimum Reflux

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Xa

Ya

XD

Min Reflux happens when the two operating lines intersect on equilibrium curve

XB

Minimum Reflux

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Xa

Ya

XD

Don’t forget the q line. Min reflux occurs at intersection with equilibrium curve because all three lines should intersect

XB

Calculation of Minimum Reflux Ratio, Rm

Based on the previous figure, the slope of the line is given by

Dnn xR

xR

Ry

1

1

11

At this point: xn=x* and yn+1=y*

**

*

1

1*

1*

min

minmin

min

xy

yxR

xR

xR

Ry

D

D

y

x

DxR 1

1

**

*

1

1*

1*

min

minmin

min

xy

yxR

xR

xR

Ry

D

D

x*

xD

slope=R/(R+1)

xBxF

y*

Feed –liquid at bubble point (saturated liquid feed) q=1

Feed –partially vapour( 0<q<1)

Feed –cold liquid (q>1) Feed –saturated vapor (q=0)

Minimum number of plates or Total Reflux

If no product is withdrawn from the still (D=0), the column is said to operate under conditions of total reflux and, as seen from equation , the top operating line has its maximum slope of unity, and coincides with the line x=y.

Total reflux

Dnn xR

xR

Ry

1

1

11

If R=L/D= ∞ then R/(R+1)=1; also L=V

nn xy 1

F=0

D=0R=L/D=∞L/V=1

B=0

1

1

nn

nn

VL

VDL

Total Reflux

All vapour is condensed and returned as liquid

Minimum number of theoretical steps

Can use Fenske equation to calculate Nmin

(Ref. Transport Process and Separation Process Principles, Geankoplis 4th ed. Page:716)

Sometimes a column is operated in total reflux at startup

av

B

B

D

D

xx

xx

Nln

1.

1log

min

MULTI COMPONENT SYSTEMS

Separation of more than two components.

Base on the relative volatility i value of each components, (light or heavy components)

A

A, B,C

A, B

CB

1 2

Key component:

-light key

-Heavy key

MULTICOMPONENT SYSTEM

For non ideal solution (hydrocarbons), the equilibrium data can be described by K factors (distribution coefficient)

“K”= ratio of mole fraction in vapor and liquid phases at equilibrium

The value of K are available from Depriester Chart.

iii xKy

i

ii x

yK

Raoult’s law (ideal system)

K (for non ideal system-dependant on T and P)

A A Ay K x

Phase equilibrium in multicomponent

For ideal solutions, the equilibrium data can be calculated from the Raoult’s and Dalton’s Law

oiii Pxp

Pp

y ii

PP

PxPx

Koi

i

oii

i

(Raoult’s Law) (Dalton’s Law)

AAAAA

A xKP

xP

P

py

Phase equilibrium in multicomponent Relative volatility (αi) for each component in a

multicomponent can be defined similar with binary mixture.

If component C in a mixture of A, B, C and D is selected as the base component,

C

ii K

K o

j

oi

ij PP

Phase equilibrium in multicomponent K factor strongly temperature dependent

because of the change in vapor pressure.

The ratio of K factor is the same as the relative volatility of components:

oj

oi

j

i

jj

iiij P

P

K

K

xy

xy

/

/

MULTICOMPONENT SYSTEM

Bubble Point ….initial boiling point of a liquid mixture.

Must satisfy the relation yi=1.0

The temperature is assumed and values of Ki are obtained from vapor pressure data and the known total pressure.

0.1iii xKy

MULTICOMPONENT SYSTEM

Bubble Point If the summation Kixi > 1.0, a lower

temperature is chosen and repeat the calculation until the equation is satisfied.

If the summation Kixi = 1.0, the composition of the vapor in equilibrium with liquid

Bubble PointFor a mixture of A, B, C and D with C as the base

component: Assume the temperature.

Calculate the value of αi from the value of Ki at this temperature.

Calculate the value of KC from

Compare the temperature corresponding to the calculated value of KC to the assumed temperature.

ii

C xK

0.1

Bubble Point If the values differ, the calculated

temperature is used for the next iteration. After the final temperature is known, the

vapor composition is calculated from

ii

iii x

xy

Example 1

A liquid feed to a distillation tower at 405.3 kPa abs is fed to a distillation tower.The composition in mole fractions is as follows:

n-butane(xA=0.40),

n-pentane(xB=0.25),

n-hexane(xC=0.20),

n-heptane(xD=0.15).

Calculate the boiling point and the vapor in equilibrium with the liquid. Let n-hexane will be the base component.

Solution: Assume a temperature and find the K values for all component.

Depriester Chart

T = 65oC

Cont’

Assuming T = 65oC

1/ 1/ 3.643 0.2745C i iK x Referring to figure 11.7-2, at 0.2745, the T is 69oC.

j

iij K

K

Get the K value from from Depriester Chart.

For the second trial, use 69oC and follow the same procedure.

MULTICOMPONENT SYSTEMDew Point ...initial condensation temperature

Must satisfy the relation xi=1.0

Also trial and error calculation After final T is known, liquid composition calculated from

ii

iii y

yx

01.

Ky

xi

ii

EXAMPLE: BOILING POINT,DEW POINT, AND FLASH

VAPORIZATION OF MULTICOMPONENT FEED

A liquid feed to a distillation tower at 405.3 Kpa abs is fed to a distillation tower. The composition in mole fractions is as follows:

N-butane (xA=0.40)

N-pentane (xB=0.25)

N-hexane (xC=0.20)----------base component

N-heptane (xD=0.15)

a) Calculate the boiling point of feed and composition of vapor in equilibrium.

b) Calculate the dew point of feed and composition of liquid in equilibrium.

SOLUTION

a) Calculate the boiling point and composition of vapor in equilibrium

1) Assume 1st temperature = 650C

2) Obtain value of K from Depriester Chart

3) Construct temperature trial table

4) Stop the iteration when the assumed temperature gives same values with the exact temperature. (means it is the bubble point)

Trial 1 (Temperature = 650C) Comp xi Ki αixi

A 0.40 1.68 6.857 2.743

B 0.25 0.63 2.571 0.643

C 0.20 0.245 1.000 0.200

D 0.15 0.093 0.380 0.057

Total 1.00 Σαixi = 3.643

Kc = 1/Σαixi = 1/3.643 = 0.2745 (690C)----get from Depriester chart

* Since the final temperature is not same with the assume temperature, do next trial using last temperature.

Trial 2 (Temperature = 690C) Comp xi Ki αixi

A 0.40

B 0.25

C 0.20

D 0.15

Total 1.00 Σαixi =

Kc = 1/Σαixi = 1/(……) = ……(….0C)----get from Depriester chart

* Since the final temperature is not same with the assume temperature, do next trial using last temperature.

Trial 3 (Temperature = 700C) Comp xi Ki αixi yi

A 0.40 1.86 6.607 2.643 0.748

B 0.25 0.710 2.522 0.631 0.178

C 0.20 0.2815 1.000 0.200 0.057

D 0.15 0.110 0.391 0.059 0.017

Total 1.00 Σαixi = 3.533 1.000

Kc = 1/Σαixi = 1/3.533 = 0.2830 (700C)----get from Depriester chart

* Since the final temperature is same with the assume temperature, stop the iteration.* The last value of temperature is called bubble point.

SOLUTION

b) Calculate the dew point and composition of liquid in equilibrium

1) Assume 1st temperature = 1060C

2) Obtain value of K from Depriester Chart

3) Construct temperature trial table

4) Stop the iteration when the assumed temperature gives same values with the exact temperature. (means it is the dew point)

Trial 1 (Temperature = 1060C) Comp yi Ki yi/αi

A 0.40 3.50 5.036 0.0794

B 0.25 1.54 2.216 0.1128

C 0.20 0.695 1.000 0.2000

D 0.15 0.330 0.4748 0.3159

Total 1.00 Σyi/αi = 0.7081

Kc = Σyi/αi = 0.7081 get temperature from Depriester chart

* Since the final temperature is not same with the assume temperature, do next trial using last temperature.

Trial 2 (Temperature = 1070C) Comp yi Ki yi/αi xi

A 0.40 3.55 4.931 0.0811 0.114

B 0.25 1.60 2.222 0.1125 0.158

C 0.20 0.720 1.000 0.2000 0.281

D 0.15 0.340 0.472 0.3178 0.447

Total 1.00 Σyi/αi =0.7114 1.000

Kc = yi/αi= 0.7114 -----Close to Kc actual at (1070C)=0.720 (get from

Depriester chart)*Kc that have been calculated is close enough with Kc at 1070C from Depriester Chart. *Thus, the last value of temperature is called dew point.

THE END